We are tasked with finding the number of five-digit numbers that can be formed using the digits 1, 2, 3, 6, 9, with the following conditions:

• The first digit must be 1.
• One of the other digits must be 2.
• No digit can appear more than once.

Step-by-step Solution:

1. Fix the first digit: Since the first digit must be 1, the number will always start with 1. So, the problem now reduces to forming a four-digit number using the digits 2, 3, 6, 9, where 2 must appear exactly once.

2. Choose the position for digit 2: Out of the remaining four positions (after fixing the first digit), we need to choose one position for the digit 2. There are 4 possible positions for placing 2.

3. Fill the remaining three positions: After placing 2, we are left with three available positions and three remaining digits (3, 6, 9). Since no digit can repeat, we can arrange the digits 3, 6, 9 in the remaining three positions in $3!$ ways (since the order matters).

$3! = 3 \times 2 \times 1 = 6$

4. Total number of possibilities: The total number of five-digit numbers is the product of the choices for the position of 2 and the number of ways to arrange the remaining three digits.

$\text{Total} = 4 \times 3! = 4 \times 6 = 24$

Thus, the number of five-digit numbers that can be formed under the given conditions is 24.

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Would you like more details on any step? Here are 5 related questions for deeper understanding:

1. What if the digit 1 could also be repeated in other positions? How would the solution change?
2. How would the problem change if there were no restrictions on which digit must appear?
3. What if the condition was that exactly two positions must contain the digit 2?
4. How can you generalize this approach to problems with different sets of digits?
5. How would the solution change if the first digit had to be even?

Tip: When dealing with problems involving the arrangement of digits, always clarify any repetition rules and special conditions before counting permutations.